Mixing
How many permutations of this alphanumerical string are possible under these conditions?
Clash Royale CLAN TAG#URR8PPP
up vote
-1
down vote
favorite
The string is in the format:
$N_1 N_2 L_1 L_2$
Under the following conditions:
$N_1$ and $N_2$ can be numbers 0-9, and $L_1$ and $L_2$ can be letters A-Z.
Initially, I thought it would just be $(99 x 26^2)-26=66,898$ but I have a feeling I'm missing something.
Thanks
combinatorics permutations
asked Jul 24 at 13:41
Phizzy
1085
add a comment |Â
up vote
-1
down vote
favorite
The string is in the format:
$N_1 N_2 L_1 L_2$
Under the following conditions:
$N_1$ and $N_2$ can be numbers 0-9, and $L_1$ and $L_2$ can be letters A-Z.
Initially, I thought it would just be $(99 x 26^2)-26=66,898$ but I have a feeling I'm missing something.
Thanks
combinatorics permutations
asked Jul 24 at 13:41
Phizzy
1085
You are missing two things: the amount of possible numbers is 100 instead of 99 (maybe you forgot 00?) and if you don't want permutations of the string repeated, you should divide by two the number of combinations of letters (you don't want AB and BA at the same time as one is a permutation of the other)
– M4g1ch
Jul 24 at 14:18
You've just made me realise my question was wrong slightly - AB and BA is allowed. Because of how I was evaluating the permutations in my head I somehow reached the conclusion that there could be two instances of AA...ZZ etc. A bit silly really. Edited to remove the duplicates statement.
– Phizzy
Jul 24 at 14:25
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
The string is in the format:
$N_1 N_2 L_1 L_2$
Under the following conditions:
$N_1$ and $N_2$ can be numbers 0-9, and $L_1$ and $L_2$ can be letters A-Z.
Initially, I thought it would just be $(99 x 26^2)-26=66,898$ but I have a feeling I'm missing something.
Thanks
combinatorics permutations
asked Jul 24 at 13:41
Phizzy
1085
The string is in the format:
$N_1 N_2 L_1 L_2$
Under the following conditions:
$N_1$ and $N_2$ can be numbers 0-9, and $L_1$ and $L_2$ can be letters A-Z.
Initially, I thought it would just be $(99 x 26^2)-26=66,898$ but I have a feeling I'm missing something.
Thanks
combinatorics permutations
asked Jul 24 at 13:41
Phizzy
1085
edited Jul 24 at 14:26
asked Jul 24 at 13:41
Phizzy
1085
asked Jul 24 at 13:41
Phizzy
1085
asked Jul 24 at 13:41
Phizzy
1085
1085
You are missing two things: the amount of possible numbers is 100 instead of 99 (maybe you forgot 00?) and if you don't want permutations of the string repeated, you should divide by two the number of combinations of letters (you don't want AB and BA at the same time as one is a permutation of the other)
– M4g1ch
Jul 24 at 14:18
You've just made me realise my question was wrong slightly - AB and BA is allowed. Because of how I was evaluating the permutations in my head I somehow reached the conclusion that there could be two instances of AA...ZZ etc. A bit silly really. Edited to remove the duplicates statement.
– Phizzy
Jul 24 at 14:25
add a comment |Â
You are missing two things: the amount of possible numbers is 100 instead of 99 (maybe you forgot 00?) and if you don't want permutations of the string repeated, you should divide by two the number of combinations of letters (you don't want AB and BA at the same time as one is a permutation of the other)
– M4g1ch
Jul 24 at 14:18
You've just made me realise my question was wrong slightly - AB and BA is allowed. Because of how I was evaluating the permutations in my head I somehow reached the conclusion that there could be two instances of AA...ZZ etc. A bit silly really. Edited to remove the duplicates statement.
– Phizzy
Jul 24 at 14:25
You are missing two things: the amount of possible numbers is 100 instead of 99 (maybe you forgot 00?) and if you don't want permutations of the string repeated, you should divide by two the number of combinations of letters (you don't want AB and BA at the same time as one is a permutation of the other)
– M4g1ch
Jul 24 at 14:18
You are missing two things: the amount of possible numbers is 100 instead of 99 (maybe you forgot 00?) and if you don't want permutations of the string repeated, you should divide by two the number of combinations of letters (you don't want AB and BA at the same time as one is a permutation of the other)
– M4g1ch
Jul 24 at 14:18
You've just made me realise my question was wrong slightly - AB and BA is allowed. Because of how I was evaluating the permutations in my head I somehow reached the conclusion that there could be two instances of AA...ZZ etc. A bit silly really. Edited to remove the duplicates statement.
– Phizzy
Jul 24 at 14:25
You've just made me realise my question was wrong slightly - AB and BA is allowed. Because of how I was evaluating the permutations in my head I somehow reached the conclusion that there could be two instances of AA...ZZ etc. A bit silly really. Edited to remove the duplicates statement.
– Phizzy
Jul 24 at 14:25
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
As the permutations of the resulting string should all be unique, then:
- Numbers can be selected in $10cdot 9$ ways
- Letters can be selected in $26cdot 25$ ways
Thus there are $10cdot 9cdot 26cdot 25 = 58500$ possible combinations.
EDIT
If there are no restrictions for permutations, then there are $10^2 26^2$ possible ways to construct the string.
There are 4 possibilities (in brackets is the number of unique permutations of the resulting string):
- Numbers are equal, letters are different ($frac4!2! = 12$)
- Numbers are equal, letters are equal ($frac4!2!2! = 6$)
- Numbers are different, letters are equal($frac4!2! = 12$)
- Numbers are different, letters are different($4! = 24$)
answered Jul 24 at 14:03
Jaroslaw Matlak
3,880830
"Duplicate letters are not allowed". That doesn't mean duplicate numbers aren't allowed
– M4g1ch
Jul 24 at 14:09
The question stated, that all permutations of the resulting string should be unique, thus the strings 00_ , 11 , ..., 99 _ are also not allowed.
– Jaroslaw Matlak
Jul 24 at 14:59
Just to aid in my understanding, assuming that the condition about the duplicates still applied, why would 00,11...99 be excluded? My statement was intended to mean that permutations of the entire string must be unique, therefore, AFAIK, 00AB and 00YG, for example, would be permitted.
– Phizzy
Jul 24 at 15:40
@Phizzy Yup. The permutations of string abcd are, amongs others: abcd, bacd. If we say, that a=b=1, c=A, d=B (11AB), then these permutations are not unique.
– Jaroslaw Matlak
Jul 25 at 7:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
As the permutations of the resulting string should all be unique, then:
- Numbers can be selected in $10cdot 9$ ways
- Letters can be selected in $26cdot 25$ ways
Thus there are $10cdot 9cdot 26cdot 25 = 58500$ possible combinations.
EDIT
If there are no restrictions for permutations, then there are $10^2 26^2$ possible ways to construct the string.
There are 4 possibilities (in brackets is the number of unique permutations of the resulting string):
- Numbers are equal, letters are different ($frac4!2! = 12$)
- Numbers are equal, letters are equal ($frac4!2!2! = 6$)
- Numbers are different, letters are equal($frac4!2! = 12$)
- Numbers are different, letters are different($4! = 24$)
answered Jul 24 at 14:03
Jaroslaw Matlak
3,880830
"Duplicate letters are not allowed". That doesn't mean duplicate numbers aren't allowed
– M4g1ch
Jul 24 at 14:09
The question stated, that all permutations of the resulting string should be unique, thus the strings 00_ , 11 , ..., 99 _ are also not allowed.
– Jaroslaw Matlak
Jul 24 at 14:59
Just to aid in my understanding, assuming that the condition about the duplicates still applied, why would 00,11...99 be excluded? My statement was intended to mean that permutations of the entire string must be unique, therefore, AFAIK, 00AB and 00YG, for example, would be permitted.
– Phizzy
Jul 24 at 15:40
@Phizzy Yup. The permutations of string abcd are, amongs others: abcd, bacd. If we say, that a=b=1, c=A, d=B (11AB), then these permutations are not unique.
– Jaroslaw Matlak
Jul 25 at 7:30
add a comment |Â
up vote
0
down vote
accepted
As the permutations of the resulting string should all be unique, then:
- Numbers can be selected in $10cdot 9$ ways
- Letters can be selected in $26cdot 25$ ways
Thus there are $10cdot 9cdot 26cdot 25 = 58500$ possible combinations.
EDIT
If there are no restrictions for permutations, then there are $10^2 26^2$ possible ways to construct the string.
There are 4 possibilities (in brackets is the number of unique permutations of the resulting string):
- Numbers are equal, letters are different ($frac4!2! = 12$)
- Numbers are equal, letters are equal ($frac4!2!2! = 6$)
- Numbers are different, letters are equal($frac4!2! = 12$)
- Numbers are different, letters are different($4! = 24$)
answered Jul 24 at 14:03
Jaroslaw Matlak
3,880830
"Duplicate letters are not allowed". That doesn't mean duplicate numbers aren't allowed
– M4g1ch
Jul 24 at 14:09
The question stated, that all permutations of the resulting string should be unique, thus the strings 00_ , 11 , ..., 99 _ are also not allowed.
– Jaroslaw Matlak
Jul 24 at 14:59
Just to aid in my understanding, assuming that the condition about the duplicates still applied, why would 00,11...99 be excluded? My statement was intended to mean that permutations of the entire string must be unique, therefore, AFAIK, 00AB and 00YG, for example, would be permitted.
– Phizzy
Jul 24 at 15:40
@Phizzy Yup. The permutations of string abcd are, amongs others: abcd, bacd. If we say, that a=b=1, c=A, d=B (11AB), then these permutations are not unique.
– Jaroslaw Matlak
Jul 25 at 7:30
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
As the permutations of the resulting string should all be unique, then:
- Numbers can be selected in $10cdot 9$ ways
- Letters can be selected in $26cdot 25$ ways
Thus there are $10cdot 9cdot 26cdot 25 = 58500$ possible combinations.
EDIT
If there are no restrictions for permutations, then there are $10^2 26^2$ possible ways to construct the string.
There are 4 possibilities (in brackets is the number of unique permutations of the resulting string):
- Numbers are equal, letters are different ($frac4!2! = 12$)
- Numbers are equal, letters are equal ($frac4!2!2! = 6$)
- Numbers are different, letters are equal($frac4!2! = 12$)
- Numbers are different, letters are different($4! = 24$)
answered Jul 24 at 14:03
Jaroslaw Matlak
3,880830
As the permutations of the resulting string should all be unique, then:
- Numbers can be selected in $10cdot 9$ ways
- Letters can be selected in $26cdot 25$ ways
Thus there are $10cdot 9cdot 26cdot 25 = 58500$ possible combinations.
EDIT
If there are no restrictions for permutations, then there are $10^2 26^2$ possible ways to construct the string.
There are 4 possibilities (in brackets is the number of unique permutations of the resulting string):
- Numbers are equal, letters are different ($frac4!2! = 12$)
- Numbers are equal, letters are equal ($frac4!2!2! = 6$)
- Numbers are different, letters are equal($frac4!2! = 12$)
- Numbers are different, letters are different($4! = 24$)
answered Jul 24 at 14:03
Jaroslaw Matlak
3,880830
edited Jul 25 at 7:41
answered Jul 24 at 14:03
Jaroslaw Matlak
3,880830
answered Jul 24 at 14:03
Jaroslaw Matlak
3,880830
answered Jul 24 at 14:03
Jaroslaw Matlak
3,880830
3,880830
"Duplicate letters are not allowed". That doesn't mean duplicate numbers aren't allowed
– M4g1ch
Jul 24 at 14:09
The question stated, that all permutations of the resulting string should be unique, thus the strings 00_ , 11 , ..., 99 _ are also not allowed.
– Jaroslaw Matlak
Jul 24 at 14:59
Just to aid in my understanding, assuming that the condition about the duplicates still applied, why would 00,11...99 be excluded? My statement was intended to mean that permutations of the entire string must be unique, therefore, AFAIK, 00AB and 00YG, for example, would be permitted.
– Phizzy
Jul 24 at 15:40
@Phizzy Yup. The permutations of string abcd are, amongs others: abcd, bacd. If we say, that a=b=1, c=A, d=B (11AB), then these permutations are not unique.
– Jaroslaw Matlak
Jul 25 at 7:30
add a comment |Â
"Duplicate letters are not allowed". That doesn't mean duplicate numbers aren't allowed
– M4g1ch
Jul 24 at 14:09
The question stated, that all permutations of the resulting string should be unique, thus the strings 00_ , 11 , ..., 99 _ are also not allowed.
– Jaroslaw Matlak
Jul 24 at 14:59
Just to aid in my understanding, assuming that the condition about the duplicates still applied, why would 00,11...99 be excluded? My statement was intended to mean that permutations of the entire string must be unique, therefore, AFAIK, 00AB and 00YG, for example, would be permitted.
– Phizzy
Jul 24 at 15:40
@Phizzy Yup. The permutations of string abcd are, amongs others: abcd, bacd. If we say, that a=b=1, c=A, d=B (11AB), then these permutations are not unique.
– Jaroslaw Matlak
Jul 25 at 7:30
"Duplicate letters are not allowed". That doesn't mean duplicate numbers aren't allowed
– M4g1ch
Jul 24 at 14:09
"Duplicate letters are not allowed". That doesn't mean duplicate numbers aren't allowed
– M4g1ch
Jul 24 at 14:09
The question stated, that all permutations of the resulting string should be unique, thus the strings 00_ , 11 , ..., 99 _ are also not allowed.
– Jaroslaw Matlak
Jul 24 at 14:59
The question stated, that all permutations of the resulting string should be unique, thus the strings 00_ , 11 , ..., 99 _ are also not allowed.
– Jaroslaw Matlak
Jul 24 at 14:59
Just to aid in my understanding, assuming that the condition about the duplicates still applied, why would 00,11...99 be excluded? My statement was intended to mean that permutations of the entire string must be unique, therefore, AFAIK, 00AB and 00YG, for example, would be permitted.
– Phizzy
Jul 24 at 15:40
Just to aid in my understanding, assuming that the condition about the duplicates still applied, why would 00,11...99 be excluded? My statement was intended to mean that permutations of the entire string must be unique, therefore, AFAIK, 00AB and 00YG, for example, would be permitted.
– Phizzy
Jul 24 at 15:40
@Phizzy Yup. The permutations of string abcd are, amongs others: abcd, bacd. If we say, that a=b=1, c=A, d=B (11AB), then these permutations are not unique.
– Jaroslaw Matlak
Jul 25 at 7:30
@Phizzy Yup. The permutations of string abcd are, amongs others: abcd, bacd. If we say, that a=b=1, c=A, d=B (11AB), then these permutations are not unique.
– Jaroslaw Matlak
Jul 25 at 7:30
add a comment |Â
Â
draft saved
draft discarded
Â
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861343%2fhow-many-permutations-of-this-alphanumerical-string-are-possible-under-these-con%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
You are missing two things: the amount of possible numbers is 100 instead of 99 (maybe you forgot 00?) and if you don't want permutations of the string repeated, you should divide by two the number of combinations of letters (you don't want AB and BA at the same time as one is a permutation of the other)
– M4g1ch
Jul 24 at 14:18
You've just made me realise my question was wrong slightly - AB and BA is allowed. Because of how I was evaluating the permutations in my head I somehow reached the conclusion that there could be two instances of AA...ZZ etc. A bit silly really. Edited to remove the duplicates statement.
– Phizzy
Jul 24 at 14:25